What is the derivative of f(t) = (-1/(t+1) , e^(2t-1) ) ?

2 Answers
Shwetank Mauria
May 7, 2017

(dy)/(dx)=(t+1)^2e^(2t-1)

Explanation:

This is a parametric form of equation i.e. f(t)=(x(t),y(t))

In such a case (dy)/(dx)=((dy)/(dt))/((dx)/(dt))

Here x(t)=-1/(t+1) and (dx)/(dt)=-(-1)/(t+1)^2=1/(t+1)^2

and y(t)=e^(2t-1) and (dy)/(dt)=2e^(2t-1)

Hence (dy)/(dx)=(2e^(2t-1))/(1/(t+1)^2)=(t+1)^2e^(2t-1)

mason m
May 7, 2017

For a parametric function f(t)=(x(t),y(t)), the function's derivative is just given by the x and y components' derivatives: f'(t)=(x'(t),y'(t)).

So:

f(t)=(-(t+1)^-1,e^(2t-1))

f'(t)=(-(-(t+1)^-2),2e^(2t-1))

f'(t)=(1/(t+1)^2,2e^(2t-1))

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